Question

A force of $(4{x^2} + 3x)\,{\text{N}}$ acts on a particle which displaces it from $x = 2\,m$ to $x = 3\,m$. The word done by the force is:
A)$32.8\,J$
B) $3.28\,J$
C) $0.328\,J$
D) Zero

Answer 1

Hint: The work done by the object is the product of the force applied to the body and the distance it covers. In our case since the force applied to the body depends on the displacement of the object itself, we will have to integrate the formula of work done from the initial to the final position.
Formula used: In this solution, we will use the following formula:
$W = \int\limits_a^b {Fdx} $ where $W$ is the work done in moving an object from position “a” to “b” under a force $F$.

Complete step by step answer:
We’ve been given that a force of $(4{x^2} + 3x)\,{\text{N}}$ acts on a particle which displaces it from $x = 2\,m$ to $x = 3\,m$. To calculate the work done, we can use the formula for work done as
$W = \int\limits_{x = 2}^{x = 3} {(4{x^2} + 3x)dx} $
On integrating the right side using the identity of integration $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $, we get
$W = \left. {\left( {\dfrac{{4{x^3}}}{3} + \dfrac{{3{x^2}}}{2}} \right)} \right|_{x = 2}^{x = 3}$
On substituting the limits, the right-hand side simplifies as
\[W = \dfrac{{4{{(3)}^3}}}{3} + \dfrac{{3{{(3)}^2}}}{2} - \dfrac{{4{{(2)}^3}}}{3} - \dfrac{{3{{(2)}^2}}}{2}\]
\[ \Rightarrow W = 32.8\,J\]

Hence the work done in moving an object from $x = 2\,m$ to $x = 3\,m$ under a force of $(4{x^2} + 3x)\,{\text{N}}$ is $32.8\,J$ which corresponds to option (A).

Note: Here we must not directly substitute the values of $x$ in the equation as the force changes with the position of the particle and hence we must integrate the product of force and distance. The work is done only depending on the initial and the final position of the object which is reflected when we integrate the equation from the initial to the final position. This implies that irrespective of the path we take to reach the final point, the work done by the object will be the same.